(再掲)>>504より en.wikipedia.org/wiki/Well-ordering_theorem Well-ordering theorem Proof from axiom of choice The well-ordering theorem follows from the axiom of choice as follows.[9] Let the set we are trying to well-order be A, and let f be a choice function for the family of non-empty subsets of A. 注）＊ For every ordinal α, define an element aα that is in A by setting aα= f(A∖{aξ∣ξ<α}) if this complement A∖{aξ∣ξ<α} is nonempty, or leave aα undefined if it is. That is, aα is chosen from the set of elements of A that have not yet been assigned a place in the ordering (or undefined if the entirety of A has been successfully enumerated). Then the order < on A defined by aα<aβ if and only if α<β (in the usual well-order of the ordinals) is a well-order of A as desired, of order type sup{α∣aα is defined}. Notes 9＾ Jech, Thomas (2002). Set Theory (Third Millennium Edition). Springer. p. 48. ISBN 978-3-540-44085-7. 注）＊ That we can do by induction, using a choice fiunction f for the family S of all nonempty subsets of A. (引用終り)
さて、この en.wikipedia Well-ordering theorem の Proof from axiom of choice by 9＾ Jech, Thomas (2002). Set Theory で ここの記載 ”For every ordinal α, define an element aα that is in A by setting aα= f(A∖{aξ∣ξ<α}) if this complement A∖{aξ∣ξ<α} is nonempty, or leave aα undefined if it is.” が、循環論法だと？ 気は確かか？ｗ
”aα= f(A∖{aξ∣ξ<α})”において 明らかに f 選択関数 で 定義域の集合族 A∖{aξ∣ξ<α} これが、関数の入力で aα が、関数 fの出力で a ∈A で aα は aが順序数αで添え字付けできたことを表す 順序を ”defined by aα<aβ if and only if α<β”とすれば aは、整列できたことになる （ここ aα<a'β とでもしておく方がいいかもね ；ｐ）
で、循環論法だと？ おれに言わずに、Jech, Thomas にお手紙書いてね 返事が来たら、ここにアップしてくれｗｗ ；ｐ） 笑える おサルさんよ>>7-10 ｗｗｗ ；ｐ）
